Transcript Chapter 6: The Standard Deviation as a Ruler and the Normal Model

Chapter 6:
The Standard Deviation as
a Ruler and the Normal
Model
By Marisa Suzuki
Standard Deviation as a Ruler
 Standard deviations can be used to compare very
different-looking data
 Standard deviation tells us how the whole
collection of values varies
 The more variability in data, the higher the
standard deviation will be
Standardizing with z-scores
 Results can be standardized, resulting in values denoted
with the letter z, but are usually called z-scores
 Z-scores measure the distance of each data value from
the mean in standard deviations
 Can be found using the equation:
z
y y
s
Standardizing with z-scores cont.
z
y y
s
 The mean of all of the data is subtracted from the
sample value y, which is then divided by the standard
deviation.
 A negative z-score tells us that the data is below the
mean, while a positive z-score tells us that the data is
above the mean
Combining Two Variables
 To find the standard deviation of the sum or
difference, you must ADD the variances and then
take the SQUARE ROOT
 a b   a   b
 a b   a   b
 a b 
2
2
a b
Normal Models
 Distributions that are unimodal and roughly symmetric
can be represented by a normal model (bell-shaped
curve)
 N(μ,σ) denotes a normal model
 µ = (mew) mean of the model
 σ = (lower case sigma) standard deviation in the model
 The standard normal model is a model with a mean of
0 and SD of 1, denoted as N(0,1)
Normal Models cont.
 Normal models give us an idea of how extreme a
value is by telling us how likely it is to find one that
far from the mean
 The 68-95-99.7 Rule
 68% of the values fall within 1 standard deviation of the
mean
 95% of the values fall within 2 standard deviations of the
mean
 99.7% of the values fall within 3 standard deviations of the
mean
The Bell-Shaped Curve
Question #35
 Based on the normal model for weights of Angus
steers N(1152,85), what are the cutoff values for:
a) The highest 10% of the weights?
b) The lowest 20% of the weights?
c) The middle 40% of the weights?
Question #35 (part a)
 The highest 10% of the weights?
2ND  DISTR  invNorm (0.9, 0, 1)  1.282
z=(y-µ)/σ
1.282=(y-1152)/84
y=1259.7 lbs
Question #35 (part b)
 The lowest 20% of the weights?
2ND  DISTR  invNorm (0.2, 0, 1)  -0.842
z=(y-µ)/σ
-0.842=(y-1152)/84
y=1081.3 lbs
Question #35 (part c)
 The middle 40% of the weights?
2ND  DISTR  invNorm (0.3, 0, 1)  -0.524
2ND  DISTR  invNorm (0.7, 0, 1)  0.524
z=(y-µ)/σ
-0.524=(y-1152)/84
0.524=(y-1152)/84
y=1108.0 lbs
y=1196.0 lbs
Question #37
 Consider the Angus steer model N(1152,84) again:
a) What weight represents the 40th percentile?
b) What weight represents the 99th percentile?
c) What’s the IQR of the weights of these Angus
steers?
Question #37 (part a)
 What weight represents the 40th percentile?
2ND  DISTR  invNorm (0.4, 0, 1)  -0.253
z=(y-µ)/σ
-0.253=(y-1152)/84
y=1130.7 lbs
Question #37 (part b)
 What weight represents the 99th percentile?
2ND  DISTR  invNorm (0.99, 0, 1)  2.326
z=(y-µ)/σ
2.326=(y-1152)/84
y=1347.4 lbs
Question #37 (part c)
 What’s the IQR of the weights of these Angus
steers?
2ND  DISTR  invNorm (0.25 0, 1)  -0.674
2ND  DISTR  invNorm (0.75, 0, 1)  0.674
Q1 = 1095.34
Q3 = 1208.60
IQR = (Q3 – Q1) = (1208.60 – 1095.34) = 113.3 lbs